Question

In a physics lab students are conducting an experiment to learn about the heat capacity of different materials. The first group is instructed to add 1.5-g lead pellets at a temperature of 92°C to 275 g of water at 16°C. A second group is given the same number of 1.5-g pellets as the first group, but these are now aluminum pellets. Assume that no heat is lost to or gained from the surroundings for either group.

(a) If the final equilibrium temperature of the lead pellets and
water is 27°C, how many whole pellets did the first group use in
the experiment? The specific heat of lead is0.0305 kcal/(kg ·
°C).

________pellets

(b) What is the equilibrium temperature of the aluminum and water
mixture for the second group?

_______ °C

Answer #1

let the number of lead pellet be n

So mass of lead = 1.5n g

initial temperature of lead = 92°C

Final temperature = 27 °C

So heat lost by lead = 1.5n * 0.0305 * (92-27) = 2.97375n calories

Mass of water = 275 g

initial temperature of water = 16 °C

Specific heat capacity of water = 1 cal/g-C

So gain of heat = 275 * 1 * (27-16) = 3025 calories

We know *heat gained by the cold water* = *heat lost
by the hot pellets*

=> 3025 = 2.97375n

=> n = 1017.23

So number of lead pellet = **1017 (approx)**

b) mass of Al = 1.5 * 1017 = 1525.5 g

specific heat of Al = 0.215 cal/g-C

let the final temperature be T

Heat lost by Al = 1525.5 * 0.215 * (92 - T)

Heat gained by water = 275 * 1 * (T -16)

So heat lost = heat gained

=> 1525.5 * 0.215 * (92 - T) = 275 * 1 * (T -16)

=> 1.19266 (92 - T) = (T - 16)

=> 2.19266 T = 125.725

=> T = 57.34 °C

So the equilibrium temperature of the aluminum and water mixture
for the second group = **57.34 °C**

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